\(\int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx\) [1196]

Optimal result

Integrand size = 26, antiderivative size = 128 \[ \int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx=\frac {(B d-A e) \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}-\frac {(b B d-2 A c d+A b e) \text {arctanh}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{2 d^{3/2} (c d-b e)^{3/2}} \]

[Out]

-1/2*(A*b*e-2*A*c*d+B*b*d)*arctanh(1/2*(b*d+(-b*e+2*c*d)*x)/d^(1/2)/(-b*e+c*d)^(1/2)/(c*x^2+b*x)^(1/2))/d^(3/2
)/(-b*e+c*d)^(3/2)+(-A*e+B*d)*(c*x^2+b*x)^(1/2)/d/(-b*e+c*d)/(e*x+d)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {820, 738, 212} \[ \int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx=\frac {\sqrt {b x+c x^2} (B d-A e)}{d (d+e x) (c d-b e)}-\frac {(A b e-2 A c d+b B d) \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{2 d^{3/2} (c d-b e)^{3/2}} \]

[In]

Int[(A + B*x)/((d + e*x)^2*Sqrt[b*x + c*x^2]),x]

[Out]

((B*d - A*e)*Sqrt[b*x + c*x^2])/(d*(c*d - b*e)*(d + e*x)) - ((b*B*d - 2*A*c*d + A*b*e)*ArcTanh[(b*d + (2*c*d -
b*e)*x)/(2*Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(2*d^(3/2)*(c*d - b*e)^(3/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Dist[
(b*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[S
implify[m + 2*p + 3], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(B d-A e) \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}-\frac {(b B d-2 A c d+A b e) \int \frac {1}{(d+e x) \sqrt {b x+c x^2}} \, dx}{2 d (c d-b e)} \\ & = \frac {(B d-A e) \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}+\frac {(b B d-2 A c d+A b e) \text {Subst}\left (\int \frac {1}{4 c d^2-4 b d e-x^2} \, dx,x,\frac {-b d-(2 c d-b e) x}{\sqrt {b x+c x^2}}\right )}{d (c d-b e)} \\ & = \frac {(B d-A e) \sqrt {b x+c x^2}}{d (c d-b e) (d+e x)}-\frac {(b B d-2 A c d+A b e) \tanh ^{-1}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{2 d^{3/2} (c d-b e)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx=\frac {\sqrt {x} \left (\frac {\sqrt {d} (B d-A e) \sqrt {x} (b+c x)}{(c d-b e) (d+e x)}-\frac {(b B d-2 A c d+A b e) \sqrt {b+c x} \arctan \left (\frac {-e \sqrt {x} \sqrt {b+c x}+\sqrt {c} (d+e x)}{\sqrt {d} \sqrt {-c d+b e}}\right )}{(-c d+b e)^{3/2}}\right )}{d^{3/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[(A + B*x)/((d + e*x)^2*Sqrt[b*x + c*x^2]),x]

[Out]

(Sqrt[x]*((Sqrt[d]*(B*d - A*e)*Sqrt[x]*(b + c*x))/((c*d - b*e)*(d + e*x)) - ((b*B*d - 2*A*c*d + A*b*e)*Sqrt[b
+ c*x]*ArcTan[(-(e*Sqrt[x]*Sqrt[b + c*x]) + Sqrt[c]*(d + e*x))/(Sqrt[d]*Sqrt[-(c*d) + b*e])])/(-(c*d) + b*e)^(
3/2)))/(d^(3/2)*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {\frac {\left (A e -B d \right ) \sqrt {x \left (c x +b \right )}}{e x +d}-\frac {\left (A b e -2 A c d +B b d \right ) \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )}{\sqrt {d \left (b e -c d \right )}}}{d \left (b e -c d \right )}\)	\(96\)
default	\(-\frac {B \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}+\frac {\left (A e -B d \right ) \left (\frac {e^{2} \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{d \left (b e -c d \right ) \left (x +\frac {d}{e}\right )}-\frac {\left (b e -2 c d \right ) e \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{2 d \left (b e -c d \right ) \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}\right )}{e^{3}}\)	\(370\)

[In]

int((B*x+A)/(e*x+d)^2/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/d/(b*e-c*d)*((A*e-B*d)*(x*(c*x+b))^(1/2)/(e*x+d)-(A*b*e-2*A*c*d+B*b*d)/(d*(b*e-c*d))^(1/2)*arctan((x*(c*x+b)
)^(1/2)/x*d/(d*(b*e-c*d))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 399, normalized size of antiderivative = 3.12 \[ \int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx=\left [-\frac {{\left (A b d e + {\left (B b - 2 \, A c\right )} d^{2} + {\left (A b e^{2} + {\left (B b - 2 \, A c\right )} d e\right )} x\right )} \sqrt {c d^{2} - b d e} \log \left (\frac {b d + {\left (2 \, c d - b e\right )} x + 2 \, \sqrt {c d^{2} - b d e} \sqrt {c x^{2} + b x}}{e x + d}\right ) - 2 \, {\left (B c d^{3} + A b d e^{2} - {\left (B b + A c\right )} d^{2} e\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (c^{2} d^{5} - 2 \, b c d^{4} e + b^{2} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}\right )} x\right )}}, -\frac {{\left (A b d e + {\left (B b - 2 \, A c\right )} d^{2} + {\left (A b e^{2} + {\left (B b - 2 \, A c\right )} d e\right )} x\right )} \sqrt {-c d^{2} + b d e} \arctan \left (-\frac {\sqrt {-c d^{2} + b d e} \sqrt {c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) - {\left (B c d^{3} + A b d e^{2} - {\left (B b + A c\right )} d^{2} e\right )} \sqrt {c x^{2} + b x}}{c^{2} d^{5} - 2 \, b c d^{4} e + b^{2} d^{3} e^{2} + {\left (c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}\right )} x}\right ] \]

[In]

integrate((B*x+A)/(e*x+d)^2/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[-1/2*((A*b*d*e + (B*b - 2*A*c)*d^2 + (A*b*e^2 + (B*b - 2*A*c)*d*e)*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d -
b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) - 2*(B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e)*sq
rt(c*x^2 + b*x))/(c^2*d^5 - 2*b*c*d^4*e + b^2*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3)*x), -((A*b*d
*e + (B*b - 2*A*c)*d^2 + (A*b*e^2 + (B*b - 2*A*c)*d*e)*x)*sqrt(-c*d^2 + b*d*e)*arctan(-sqrt(-c*d^2 + b*d*e)*sq
rt(c*x^2 + b*x)/((c*d - b*e)*x)) - (B*c*d^3 + A*b*d*e^2 - (B*b + A*c)*d^2*e)*sqrt(c*x^2 + b*x))/(c^2*d^5 - 2*b
*c*d^4*e + b^2*d^3*e^2 + (c^2*d^4*e - 2*b*c*d^3*e^2 + b^2*d^2*e^3)*x)]

Sympy [F]

\[ \int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx=\int \frac {A + B x}{\sqrt {x \left (b + c x\right )} \left (d + e x\right )^{2}}\, dx \]

[In]

integrate((B*x+A)/(e*x+d)**2/(c*x**2+b*x)**(1/2),x)

[Out]

Integral((A + B*x)/(sqrt(x*(b + c*x))*(d + e*x)**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((B*x+A)/(e*x+d)^2/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 519 vs. \(2 (112) = 224\).

Time = 0.53 (sec) , antiderivative size = 519, normalized size of antiderivative = 4.05 \[ \int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx=-\frac {\frac {{\left (B b d e^{3} \log \left ({\left | 2 \, c d e - b e^{2} - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c} {\left | e \right |} \right |}\right ) - 2 \, A c d e^{3} \log \left ({\left | 2 \, c d e - b e^{2} - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c} {\left | e \right |} \right |}\right ) + A b e^{4} \log \left ({\left | 2 \, c d e - b e^{2} - 2 \, \sqrt {c d^{2} - b d e} \sqrt {c} {\left | e \right |} \right |}\right ) + 2 \, \sqrt {c d^{2} - b d e} B \sqrt {c} d e {\left | e \right |} - 2 \, \sqrt {c d^{2} - b d e} A \sqrt {c} e^{2} {\left | e \right |}\right )} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{\sqrt {c d^{2} - b d e} c d^{2} {\left | e \right |} - \sqrt {c d^{2} - b d e} b d e {\left | e \right |}} - \frac {2 \, {\left (B d e \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - A e^{2} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )\right )} \sqrt {c - \frac {2 \, c d}{e x + d} + \frac {c d^{2}}{{\left (e x + d\right )}^{2}} + \frac {b e}{e x + d} - \frac {b d e}{{\left (e x + d\right )}^{2}}}}{c d^{2} \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{2} \mathrm {sgn}\left (e\right )^{2} - b d e \mathrm {sgn}\left (\frac {1}{e x + d}\right )^{2} \mathrm {sgn}\left (e\right )^{2}} - \frac {{\left (B b d e^{3} - 2 \, A c d e^{3} + A b e^{4}\right )} \log \left ({\left | 2 \, c d e - b e^{2} - 2 \, \sqrt {c d^{2} - b d e} {\left (\sqrt {c - \frac {2 \, c d}{e x + d} + \frac {c d^{2}}{{\left (e x + d\right )}^{2}} + \frac {b e}{e x + d} - \frac {b d e}{{\left (e x + d\right )}^{2}}} + \frac {\sqrt {c d^{2} e^{2} - b d e^{3}}}{{\left (e x + d\right )} e}\right )} {\left | e \right |} \right |}\right )}{{\left (c d^{2} - b d e\right )}^{\frac {3}{2}} {\left | e \right |} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}}{2 \, e^{2}} \]

[In]

integrate((B*x+A)/(e*x+d)^2/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-1/2*((B*b*d*e^3*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*sqrt(c)*abs(e))) - 2*A*c*d*e^3*log(abs(2*c*d*
e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*sqrt(c)*abs(e))) + A*b*e^4*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*s
qrt(c)*abs(e))) + 2*sqrt(c*d^2 - b*d*e)*B*sqrt(c)*d*e*abs(e) - 2*sqrt(c*d^2 - b*d*e)*A*sqrt(c)*e^2*abs(e))*sgn
(1/(e*x + d))*sgn(e)/(sqrt(c*d^2 - b*d*e)*c*d^2*abs(e) - sqrt(c*d^2 - b*d*e)*b*d*e*abs(e)) - 2*(B*d*e*sgn(1/(e
*x + d))*sgn(e) - A*e^2*sgn(1/(e*x + d))*sgn(e))*sqrt(c - 2*c*d/(e*x + d) + c*d^2/(e*x + d)^2 + b*e/(e*x + d)
- b*d*e/(e*x + d)^2)/(c*d^2*sgn(1/(e*x + d))^2*sgn(e)^2 - b*d*e*sgn(1/(e*x + d))^2*sgn(e)^2) - (B*b*d*e^3 - 2*
A*c*d*e^3 + A*b*e^4)*log(abs(2*c*d*e - b*e^2 - 2*sqrt(c*d^2 - b*d*e)*(sqrt(c - 2*c*d/(e*x + d) + c*d^2/(e*x +
d)^2 + b*e/(e*x + d) - b*d*e/(e*x + d)^2) + sqrt(c*d^2*e^2 - b*d*e^3)/((e*x + d)*e))*abs(e)))/((c*d^2 - b*d*e)
^(3/2)*abs(e)*sgn(1/(e*x + d))*sgn(e)))/e^2

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{(d+e x)^2 \sqrt {b x+c x^2}} \, dx=\int \frac {A+B\,x}{\sqrt {c\,x^2+b\,x}\,{\left (d+e\,x\right )}^2} \,d x \]

[In]

int((A + B*x)/((b*x + c*x^2)^(1/2)*(d + e*x)^2),x)

[Out]

int((A + B*x)/((b*x + c*x^2)^(1/2)*(d + e*x)^2), x)